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3 years ago

What is the midpoint of FB

Mathematics

1 answer:

VMariaS [17]3 years ago

3 0

Answer:

Point L as it is the midpoint of FB

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HURRY I NEED HELP: Write an equation in point-slope form of the line that passes through the point (3, 5) and has a slope of m=−

skelet666 [1.2K]

Answer:

$\huge y - 5 = - 1(x - 3) \\$

Step-by-step explanation:

To find an equation of a line in point slope form when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

From the question we have the final answer as

$y - 5 = - 1(x - 3)$

Hope this helps you

6 0

3 years ago

Please solve it CORRECTLY step by step asap

Lorico [155]

Answer:

$Product = \frac{3}{5}ab$

Step-by-step explanation:

Given

$-\frac{3}{7}a$

$-\frac{21}{15}b$

Required

The product

The product is:

$Product = -\frac{3}{7}a * -\frac{21}{15}b$ '

Cancel out the negative signs

$Product = \frac{3}{7}a * \frac{21}{15}b$

Simplify fractions

$Product = \frac{1}{7}a * \frac{21}{5}b$

$Product = \frac{1}{1}a * \frac{3}{5}b$

$Product = \frac{3}{5}ab$

6 0

3 years ago

PLZ HELP ME ASAPPPPP

vovangra [49]

AnsweP is eaqul to Q

Step-by-step explanation:

7 0

3 years ago

Can a remainder in a division problem equal the divisor

Rufina [12.5K]

Well the answer is No.A remainder will always be smaller than the divisor ,because the divisor is the one that used to divide the dividend and the remaining number that cannot be divided by the divisor is called the remainder.
Questions?

6 0

3 years ago

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0

3 years ago